7.3 FINDING THE EIGENVECTORS OF A MATRIX After we have found an eigenvalue ¸ of an n£n matrix A, we have to find the vectors ~v in Rn such that A~v = ¸~v or (¸In ¡ A)~v = ~0 In other words, we have to find the kernel of the matrix ¸In ¡ A. Definition 7.3.1 Eigenspace Consider an eigenvalue ¸ of an n £ n matrix A.Then the kernel of the matrix ¸In¡A is called the eigenspace associated with ¸, denoted by E¸: E¸ = ker(¸In ¡ A) Note that E¸ consists of all solutions ~v of the linear system A~v = ¸~v 1 EXAMPLE 1 Let T (~x) = A~v be the orthogo- nal projection onto a plane E in R3. Describe the eigenspaces geometrically. Solution See Figure 1. The nonzero vectors ~v in E are eigenvectors with eigenvalue 1. Therefore, the eigenspace E1 is just the plane E. Likewise, E0 is simply the kernel of A (A~v = ~0); that is, the line E? perpendicular to E. 2 EXAMPLE" 2 Find# the eigenvectors of the 1 2 matrix A = . 4 3 Solution See Section 7.2, Example 1, we saw the eigen- values are 5 and -1. Then " # 4 ¡2 E = ker(5I ¡ A) = ker 5 2 ¡4 2 " # " # " # 4 ¡2 2 1 = ker = span = span 0 0 4 2 " # ¡2 ¡2 E = ker(¡I ¡ A) = ker ¡1 2 ¡4 ¡4 " # " # 2 1 = span = span ¡2 ¡1 Both eigenspaces are lines, See Figure 2. 3 EXAMPLE 3 Find the eigenvectors of 2 3 1 1 1 6 7 A = 4 0 0 1 5 : 0 0 1 Solution Since 2 fA(¸) = ¸(¸ ¡ 1) the eigenvalues are 1 and 0 with algebraic mul- tiplicities 2 and 1. 2 3 0 ¡1 ¡1 6 7 E1 = ker 4 0 1 ¡1 5 0 0 0 To find this kernel, apply Gauss-Jordan Elimi- nation: 2 3 2 3 0 ¡1 ¡1 0 1 1 6 7 6 7 4 0 1 ¡1 5 rref 4 0 0 1 5 ¡¡! 0 0 0 0 0 0 The general solution of the system ¯ ¯ ¯ ¯ ¯ x = 0 ¯ ¯ 2 ¯ ¯ x3 = 0 ¯ 4 is 2 3 2 3 x1 1 6 7 6 7 4 0 5 = x1 4 0 5 0 0 Therefore, 2 3 1 6 7 E1 = span 4 0 5 0 Likewise, compute the E0: 2 3 1 6 7 E0 = span 4 ¡1 5 0 Both eigenspaces are lines in the x1-x2-plane, as shown in Figure 3. Compare with Example 1. There, too, we have two eigenvalues 1 and 0, but one of the eigenspace, E1, is a plane. Definition 7.3.2 Geometric multiplicity Consider an eigenvalue ¸ if a matrix A. Then the dimension of eigenvalue E¸ = ker(¸In ¡ A) is called the geometric multiplicity of eigenvalue ¸. Thus, the geometric multiplicity of ¸ is the nullity of matrix ¸In ¡ A. Example 3 shows that the geometric multiplic- ity of an eigenvalue may be different from the algebraic multiplicity. We have (algebraic multiplicity of eigenvalue 1)=2, but (geometric multiplicity of eigenvalue 1)=1. Fact 7.3.3 Consider an eigenvalue ¸ of a matrix A. Then (geometric multiplicity of ¸)· (algebraic multiplicity of ¸). 5 EXAMPLE 4 Consider an upper triangular matrix of the form 2 3 1 ² ² ² ² 6 7 6 7 6 0 2 ² ² ² 7 6 7 A = 6 0 0 4 ² ² 7 : 6 7 4 0 0 0 4 ² 5 0 0 0 0 4 What can you say about the geometric multi- plicity of the eigenvalue 4? Solution 2 3 2 3 3 ² ² ² ² 1 ² ² ² ² 6 7 6 7 6 7 6 7 6 0 2 ² ² ² 7 6 0 1 ² ² ² 7 6 7 6 7 E4 = 6 0 0 0 ² ² 7 rref 6 0 0 0 ] ² 7 6 7 ¡¡! 6 7 4 0 0 0 0 ² 5 4 0 0 0 0 ] 5 0 0 0 0 0 0 0 0 0 0 The bullets on row 3 and 4 could be leading 1’s. Therefore, the rank of this matrix will be between 2 and 4, and its nullity will be between 3 and 1. We can conclude that the geometric multiplicity of the eigenvalue 4 is less than the algebraic multiplicity. 6 Recall Fact 7.1.3, such a basis deserves a name. Definition 7.3.4 Eigenbasis Consider an n£ n matrix A. A basis of Rn consisting of eigenvectors of A is called an eigenbasis for A. Example 1 Revisited: Projection on a plane 3 E in R . Pick a basis ~v1; ~v2 of E and a nonzero ? ~v3 in E . The vectors ~v1; ~v2; ~v3 form an eigen- basis. See Figure 4. " # 1 2 Example 2 Revisited: A = . 4 3 " # " # 1 1 The vectors and form an eigenba- 2 ¡1 sis for A, see Figure 5. 2 3 1 1 1 6 7 Example 3 Revisited: A = 4 0 0 1 5. 0 0 1 There are not enough eigenvectors to form an eigenbasis. See Figure 6. 7 EXAMPLE 5 Consider a 3 £ 3 matrix A with three eigenvalues, 1, 2, and 3. Let ~v1; ~v2, and ~v3 be corresponding eigenvectors. Are vectors ~v1; ~v2, and ~v3 necessarily linearly independent? Solution See Figure 7. Consider the plane E spanned by ~v1, and ~v2. We have to examine ~v3 can not be contained in this plane. Consider a vector ~x = c1 ~v1 + c2 ~v2 in E (with c1 6= 0 and c2 6= 0). Then A~x = c1A~v1 + c2A~v2 = c1 ~v1 + 2c2 ~v2. This vector can not be a scalar multiple of ~x; that is, E does not contain any eigenvectors besides the multiples of ~v1 and ~v2; in particular, ~v3 is not contained in E. 8 Fact 7.3.5 Considers the eigenvectors ~v1, ~v2, :::, ~vm of an n£n matrix A, with distinct eigen- values ¸1, ¸2, :::, ¸m. Then the ~vi are linearly independent. Proof We argue by induction on m. Assume the claim holds for m ¡ 1. Consider a relation c1~v1 + ¢ ¢ ¢ + cm¡1~vm¡1 + cm~vm = ~0 ² apply the transformation A to both sides: c1¸1~v1 + ¢ ¢ ¢ + cm¡1¸m¡1~vm¡1 + cm¸m~vm = ~0 ² multiply both sides by ¸m: c1¸m~v1 + ¢ ¢ ¢ + cm¡1¸m~vm¡1 + cm¸m~vm = ~0 Subtract the above two equations: c1(¸1 ¡¸m)~v1 +¢ ¢ ¢+cm¡1(¸m¡1 ¡¸m)~vm¡1 = ~0 Since ~v1, ~v2, :::, ~vm¡1 are linearly independent by induction, ci(¸i¡¸m) = 0, for i = 1; :::; m¡1. The eigenvalues are assumed to be distinct; therefore ¸i ¡ ¸m 6= 0, and ci = 0. The first equation tells us that cm~vm = ~0, so that cm = 0 as well. 9 Fact 7.3.6 If an n £ n matrix A has n distinct eigenvalues, then there is an eigenbasis for A. We can construct an eigenbasis by choosing an eigenvector for each eigenvalue. EXAMPLE 6 Is there an eigenbasis for the following matrix? 2 3 1 2 3 4 5 6 6 7 6 0 2 3 4 5 6 7 6 7 6 7 6 0 0 3 4 5 6 7 A = 6 7 6 0 0 0 4 5 6 7 6 7 4 0 0 0 0 5 6 5 0 0 0 0 0 6 Fact 7.3.7 Consider an n £ n matrix A. If the geometric multiplicities of the eigenvalues of A add up to n, then there is an eigenbasis for A: We can construct an eigenbasis by choosing a basis of each eigenspace and combining these vectors. 10 Proof Suppose the eigenvalues are ¸1, ¸2, ..., ¸m, with dim(E¸i)=di. We first choose a basis ~v1, ~v2, ..., ~vd1 of E¸1, and then a basis ~vd1+1, ..., ~vd1+d2 of E¸2, and so on. Consider a relation c ~v + ¢ ¢ ¢ + c ~v + ¢ ¢ ¢ + c ~v + ¢ ¢ ¢+¢ ¢ ¢ + cn~vn = ~0 |1 1 {z d1 d}1 | d{z1+d2 d1+d}2 | {z } ~w1 in E¸1 ~w2 in E¸2 ~wm in E¸m Each under-braced sum ~wi must be a zero vec- tor since if they are nonzero eigenvectors, they must be linearly independent and the relation can not hold. Because ~w1 = 0, it follows that c1 = c2 = ¢ ¢ ¢ = cd1 = 0, since ~v1, ~v2, ..., ~vd1 are linearly inde- pendent. Likewise, all the other cj are zero. EXAMPLE 7 Consider an Albanian mountain farmer who raises goats. This particular breed of goats has a life span of three years. At the end of each year t, the farmer conducts a census of his goats. He counts the number of young goats j(t) (those born in the year t), the middle-aged ones m(t) (born the year before), and the old ones a(t) (born in the year t ¡ 2). The state of the herd can be represented by the vector 2 3 j(t) 6 7 ~x(t) = 4 m(t) 5 a(t) How do we expect the population to change from year to year? Suppose that for this breed and environment the evolution of the system can be modelled by ~x(t + 1) = A~x(t) 2 3 0 0:95 0:6 6 7 where A = 4 0:8 0 0 5 0 0:5 0 11 We leave it as an exercise to interpret the en- tries of A in terms of reproduction rates and survival rates. Suppose the initial populations are j0 = 750 and m0 = a0 = 200.What will the popula- tions be after t years, according to this model? What will happen in the long term? Solution Step 1: Find eigenvalues. Step 2: Find eigenvectors.